二阶偏导数（Second-order Partial Derivatives） 是多元函数微分学中的重要概念。对于多元函数 $z = f(x,y)$，其一阶偏导数 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$ 仍然是 $x$ 和 $y$ 的函数，我们可以对这些偏导数再求偏导，得到二阶偏导数。

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基本定义

对于函数 $z = f(x,y)$，其二阶偏导数包括：

1. 关于 $x$ 的二阶偏导数：

   $$\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)$$

2. 关于 $y$ 的二阶偏导数：

   $$\frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial y}\right)$$

3. 混合偏导数：

   $$\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)$$

   $$\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)$$

混合偏导数的连续性定理

如果函数 $f(x,y)$ 的二阶混合偏导数 $\frac{\partial^2 f}{\partial x \partial y}$ 和 $\frac{\partial^2 f}{\partial y \partial x}$ 在某区域内连续，则在该区域内：

$$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$$

这个性质在实际计算中非常有用，可以大大简化计算过程。

复合函数的二阶偏导数

对于复合函数 $z = f(u(x,y), v(x,y))$，计算二阶偏导数需要使用链式法则。设 $u = u(x,y)$，$v = v(x,y)$，则：

$$\begin{aligned} \frac{\partial z}{\partial x} &= \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} \\ &= f_1 \cdot u_x + f_2 \cdot v_x \end{aligned}$$

其中 $f_1 = \frac{\partial f}{\partial u}$，$f_2 = \frac{\partial f}{\partial v}$。

例题1

设 $z = f(x^2 - y^2, xy)$，且 $f(u,v)$ 有连续的二阶偏导数，求 $\frac{\partial^2 z}{\partial x \partial y}$。

解：

设 $u = x^2 - y^2$，$v = xy$，则 $z = f(u,v)$。

第一步：求一阶偏导数

$$\begin{aligned} \frac{\partial z}{\partial x} &= \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} \\ &= f_1 \cdot 2x + f_2 \cdot y = 2xf_1 + yf_2 \\ \\ \frac{\partial z}{\partial y} &= \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y} \\ &= f_1 \cdot (-2y) + f_2 \cdot x = -2yf_1 + xf_2 \end{aligned}$$

第二步：求混合偏导数 $\frac{\partial^2 z}{\partial x \partial y}$

对 $\frac{\partial z}{\partial x} = 2xf_1 + yf_2$ 关于 $y$ 求偏导：

$$\begin{aligned} \frac{\partial^2 z}{\partial x \partial y} &= \frac{\partial}{\partial y}(2xf_1 + yf_2) \\ &= 2x \cdot \frac{\partial f_1}{\partial y} + f_2 + y \cdot \frac{\partial f_2}{\partial y} \end{aligned}$$

其中：

$$\begin{aligned} \frac{\partial f_1}{\partial y} &= \frac{\partial f_1}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial f_1}{\partial v} \cdot \frac{\partial v}{\partial y} \\ &= f_{11} \cdot (-2y) + f_{12} \cdot x = -2yf_{11} + xf_{12} \\ \\ \frac{\partial f_2}{\partial y} &= \frac{\partial f_2}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial f_2}{\partial v} \cdot \frac{\partial v}{\partial y} \\ &= f_{21} \cdot (-2y) + f_{22} \cdot x = -2yf_{21} + xf_{22} \end{aligned}$$

由于 $f$ 有连续的二阶偏导数，所以 $f_{12} = f_{21}$。

第三步：合并结果

$$\begin{aligned} \frac{\partial^2 z}{\partial x \partial y} &= 2x(-2yf_{11} + xf_{12}) + f_2 + y(-2yf_{12} + xf_{22}) \\ &= -4xyf_{11} + 2x^2f_{12} + f_2 - 2y^2f_{12} + xyf_{22} \\ &= f_2 - 4xyf_{11} + 2(x^2 - y^2)f_{12} + xyf_{22} \end{aligned}$$

提示

选择填空小技巧：

$$\frac{\partial^{2}z}{\partial x\partial y}=(\frac{\partial z}{\partial x})_{y}+\frac{\partial z}{\partial x}\cdot\frac{\partial z}{\partial y}$$

$$\begin{aligned} I &= (2xf_{1}+yf_{2})_{y}+(2xf_{1}+yf_{2})\cdot(-2yf_{1}+xf_{2}) \\ &= f_2-4xyf_{11}+2(x^2-y^2)f_{12}+xyf_{22} \end{aligned}$$

例题2

设 $z = f\left(\frac{y}{x}, \frac{x}{y}\right)$，其中 $f$ 有连续的二阶偏导数，求 $\frac{\partial^2 z}{\partial x^2}$。

解：

设 $u = \frac{y}{x}$，$v = \frac{x}{y}$，则 $z = f(u,v)$。

第一步：求一阶偏导数

$$\begin{aligned} \frac{\partial z}{\partial x} &= \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} \\ &= f_1 \cdot \left(-\frac{y}{x^2}\right) + f_2 \cdot \frac{1}{y} \\ &= -\frac{y}{x^2}f_1 + \frac{1}{y}f_2 \end{aligned}$$

第二步：求二阶偏导数

$$\begin{aligned} \frac{\partial^2 z}{\partial x^2} &= \frac{\partial}{\partial x}\left(-\frac{y}{x^2}f_1 + \frac{1}{y}f_2\right) \\ &= \frac{2y}{x^3}f_1 - \frac{y}{x^2} \cdot \frac{\partial f_1}{\partial x} + \frac{1}{y} \cdot \frac{\partial f_2}{\partial x} \end{aligned}$$

其中：

$$\begin{aligned} \frac{\partial f_1}{\partial x} &= \frac{\partial f_1}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f_1}{\partial v} \cdot \frac{\partial v}{\partial x} \\ &= f_{11} \cdot \left(-\frac{y}{x^2}\right) + f_{12} \cdot \frac{1}{y} \\ &= -\frac{y}{x^2}f_{11} + \frac{1}{y}f_{12} \\ \\ \frac{\partial f_2}{\partial x} &= \frac{\partial f_2}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f_2}{\partial v} \cdot \frac{\partial v}{\partial x} \\ &= f_{21} \cdot \left(-\frac{y}{x^2}\right) + f_{22} \cdot \frac{1}{y} \\ &= -\frac{y}{x^2}f_{21} + \frac{1}{y}f_{22} \end{aligned}$$

第三步：合并结果

$$\begin{aligned} \frac{\partial^2 z}{\partial x^2} &= \frac{2y}{x^3}f_1 - \frac{y}{x^2}\left(-\frac{y}{x^2}f_{11} + \frac{1}{y}f_{12}\right) + \frac{1}{y}\left(-\frac{y}{x^2}f_{21} + \frac{1}{y}f_{22}\right) \\ &= \frac{2y}{x^3}f_1 + \frac{y^2}{x^4}f_{11} - \frac{1}{x^2}f_{12} - \frac{1}{x^2}f_{21} + \frac{1}{y^2}f_{22} \\ &= \frac{2y}{x^3}f_1 + \frac{y^2}{x^4}f_{11} - \frac{2}{x^2}f_{12} + \frac{1}{y^2}f_{22} \end{aligned}$$

（这里使用了 $f_{12} = f_{21}$）